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$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$

My attempt:

the factor in the denominator implies

$$x^{2}-36=x^{2}-6^{2}$$

substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$

$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$

$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$

using trig identities: $$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$

now using integration by parts: $$\frac{1}{6}\int \sin^{-1}\theta$$ $$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$ using $uv-\int{vdu}$

$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$

now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$

it is apparent that

$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$

$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$

$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$

I have the following questions:

1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?

2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?

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  • $\begingroup$ There is a tiny mistake right after you do your substitution: you should be getting $6^3tan^3(\theta)$ at the denominator. $\endgroup$ Commented Feb 2, 2016 at 23:01
  • $\begingroup$ okay, ill try to edit appropriately $\endgroup$
    – helpmeh
    Commented Feb 2, 2016 at 23:13
  • $\begingroup$ After the correction it is very quick. Same strategy (sines and cosines). $\endgroup$ Commented Feb 2, 2016 at 23:15
  • 1
    $\begingroup$ Note that arcsin(x) and $\frac{1}{sinx}$ are different! $\endgroup$
    – Nikunj
    Commented Feb 2, 2016 at 23:16
  • $\begingroup$ I would have used $x=6\cosh(u)$ as first variable change and then $ t=e^u$ $\endgroup$
    – stity
    Commented Feb 2, 2016 at 23:18

4 Answers 4

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Be careful: with the substitution you have $$ (x^2-36)^{3/2}=(36\tan^2\theta)^{3/2}=216\tan^3\theta $$ (at least in an interval where $\tan\theta$ is positive) so your integral becomes $$ \int\frac{6\tan\theta\sec\theta}{216\tan^3\theta}\,d\theta= \frac{1}{36}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta= -\frac{1}{36}\frac{1}{\sin\theta}+C $$

If instead you set $x=6\cosh u$, you get $dx=6\sinh u$ and the identity $\cosh^2u-1=\sinh^2u$ brings the integral in the form $$ \frac{1}{36}\int\frac{1}{\sinh^2u}\,du= -\frac{1}{36}\frac{\cosh u}{\sinh u}+C= -\frac{1}{36}\frac{x}{\sqrt{x^2-36}} $$

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Without substitution:

$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{(x^2-(x^2-36))\,dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{x^2\,dx}{(x^{2}-36)^{3/2}}-\frac1{36}\int\frac{dx}{(x^{2}-36)^{1/2}}.$$

Then by parts on the first term,

$$\int\frac{x\cdot x\,dx}{(x^{2}-36)^{3/2}}=-\frac x{(x^{2}-36)^{1/2}}+\int\frac{dx}{(x^{2}-36)^{1/2}}.$$

After scaling the variable with a factor $6$, you recognize a known derivative,

$$\int\frac{dx}{\sqrt{x^2-1}}=\text{arcosh}(x).$$

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attempt 2:

$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$

the factor in the denominator implies

$$(x^{2}-36)^{3/2}=({x^{2}-6^{2}})^{3/2}$$

substituting $x=6sec\theta$, noting that $dx=6tan\theta sec\theta d\theta$

$$(x^{2}-6^{2})^{3/2}=(6^{2}sec^{2}\theta-6^{2})^{3/2}d\theta=(36tan^{2}\theta)^{3/2}=216tan^3\theta$$

$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6tan\theta sec\theta}{216tan^{3}\theta}d\theta=\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta$$

using trig identities: $$\frac{1}{36}\int\frac{sec\theta}{tan^{2}\theta}d\theta=\frac{1}{36}\int{cot(\theta)csc(\theta) d\theta}$$

the integral cot(x)csc(x) is know

$$\int cot(x)sec(x)dx=-csc(x)+C$$

so

$$\frac{1}{36}\int cot(\theta)csc(\theta)d\theta=\frac{1}{36}\bigg(-csc(\theta)\bigg)+C=-\frac{1}{36}csc(\theta)+C$$

Question 2 still applies:

2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?

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  • $\begingroup$ great, csc(x)=1/sin(x) so I have the same answer as egreg. now, is $\frac{1}{sin(x)}$ equivalent to $\frac{cosh(x)}{sinh(x)}$? $\endgroup$
    – helpmeh
    Commented Feb 2, 2016 at 23:53
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Your solution is incorrect as $$\frac{\sec\theta}{\tan\theta}$$ is not the same as $$\ sin^{-1}\theta$$ It is however, equal to $\csc\theta$, a standard integral that you had got after a few steps

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